NEET2026ChemistryChapterActual
The given P-V graph illustrates the reversible isothermal expansion of a fixed amount of an ideal gas from an initial volume V_i to a final volume V_f at two different constant temperatures, T₁ and T₂ . Let W₁ and W₂ be the magnitudes of the work done by the gas during the expansion at temperatures T₁ and T₂ , respectively. Based on the graph, which of the following statements is correct?
Options
- AW₂ > W₁ because the area under the P-V curve at T₂ is greater than the area under the curve at T₁ .
- BW₁ > W₂ because the gas exerts less pressure at lower temperatures, allowing for easier expansion.
- CW₁ = W₂ because the change in volume ( V = V_f - V_i ) is identical for both processes.
- DW₁ = W₂ because the change in internal energy ( U ) is zero for any isothermal process of an ideal gas.
Correct answer
A. W₂ > W₁ because the area under the P-V curve at T₂ is greater than the area under the curve at T₁ .
Step-by-step solution
The work done by a gas during a reversible expansion is given by the area under the P-V curve, which is mathematically represented as W = _ V_i ^ V_f P , dV . From the given graph, it is evident that the isotherm for temperature T₂ lies above the isotherm for temperature T₁ . This means that for any given volume between V_i and V_f , the pressure at T₂ is greater than the pressure at T₁ . Consequently, the area under the P-V curve at T₂ is greater than the area under the curve at T₁ . Therefore, the magnitude of th